Active Figure Charging a Capacitor The animation below shows a circuit diagram f

Question

Active Figure Charging a Capacitor The animation below shows a circuit diagram for a capacitor in series with a resistor, battery and switch (initially off). Closing the switch causes the battery to charge the capacitor. Plots of charge and current vs. time illustrate the exponential time dependence of this circuit. The animation stops at t 12.5 s Instructions Use the sliders to adjust the resistance and capacitance to see the effect on the charging of the capacitor. Click switch S to the "on" position to start the animation Explore As the capacitor charges, the circuit parameters that change do so exponentially. The rate of change is determined by the capacitive time Exercise resistanceR In each case below, calculate the time constant, then find values for the volt when t 5.00 s. Part (a) uses resistance and capacitance values that can be set with the sliders, so you can use the animation to verify your calculations. Part (b) uses values outside the slider 1.0 MS2 age across the capacitor and voltage across t he resistor =RC 1.0 2.0 | DTI a:A | It is instructive to consider how the voltages across the resistor and capacitor change in time as the capacitor charges. As the charge on the positive plate of the capacitor increases the voltage across the capacitor capacitanceC 0.03Cranges. In each case, the battery voltage is 10 V (a) Let R-1.1 M and C-1 F. 10 velts Find VR and VC when t = 5.00 s. However, as the capacitor charges, the current in the circuit decreases, finally reaching zero as the capacitor becomes fully charged. According to Ohm's law, the voltage across the resistor is proportional to the current. (b) Let R-2.9 M and C-3 f fs (reset Find VR and Vc when t- 5.00 s Display in a New Window

Explanation / Answer

a)

TIme Constant

T=RC=(1.1*106)(1*10-6) =1.1 s

Voltage across Resistor

VR=Ee-t/T =10e-5/1.1=0.106 Volts

Voltage across Capacitor

VC=E(1-e-t/T)=10(1-e-5/1.1)=9.894 Volts

b)

T=(2.9*106)(3*10-6) =8.7 s

VR=10e-5/8.7=5.63 Volts

VC=10(1-e-5/8.7) =4.37 Volts

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